Let $X_k$ be independent identically distrbuted random variables such that $$ P(X_k=1)=P(X_k=-1)=\frac{1}{2}$$ for all $k\geq 1$. Prove that $$ \sqrt{\frac{2}{n^3}}\sum_{k=1}^n kX_k$$ converges to $N(0,1)$ in distribution.
Here's what I got so far:
The characteristic function for the $X_k$ is $$\phi_{X_k}(t) = \frac{e^{it}+e^{-it}}{2} = \cos(t).$$ So, the characteristic function of $Y_n:=\sum_{k=1}^n kX_k$ is $$\phi_{Y_n}(t) = \cos(t)\cos(2t)\cdots \cos(nt).$$
So it boils down to showing that $$\lim_{n\to\infty}\phi_{ \sqrt{\frac{2}{n^3}}\sum_{k=1}^n kX_k}(t) = \lim_{n\to\infty}\prod_{k=1}^n \cos\left( \sqrt{\frac{2}{n^3}}kt\right) $$ is the characteristic function of $N(0,1)$ which is $e^{-t^2/2}$. Is the product known to be that? Or is there an easier way of doing this?